Automata & Formal Languages, Feodor F. Dragan, Kent State University 1 CHAPTER 5 Reducibility Contents Undecidable Problems from Language Theory.

Automata & Formal Languages, Feodor F. Dragan, Kent State University 2 In this chapter we examine several additional unsolvable problems. In doing so we introduce the primary method for proving that problems are computationally unsolvable. It is called reducibility. A reduction is a way of converting one problem into another problem in such a way that a solution to the second problem can be used to solve the first problem. When A is reducible to B, solving A cannot be harder that solving B because a solution to B gives a solution to A. In terms of computability theory, if A is reducible to B and B is decidable then A also is decidable. Equivalently, if A is undecidable and reducible to B, B is undecidable. This is the key to proving that various problems are undecidable. Our method for proving that a problem is undecidable will be: show that some other problem already known to be undecidable reduces to it. We will consider the following problems (~ as membership in languages): Reducibility

Automata & Formal Languages, Feodor F. Dragan, Kent State University 3 We have seen that the acceptance problem for TMs is undecidable Consider the problem determining whether a Turing machine halts (by accepting or rejecting) on a given input. S = “on input, where M is a TM and w is a string: 1.Run TM R on input. 2.If R rejects, reject. 3.If R accepts, simulate M on w until it halts. 4.If M has accepted, accept; if M rejected, reject.” The Halting Problem for TMs. Theorem 5.1: is undecidable. We use undecidability of to prove the undecidability of by reducing to. Let assume that TM R decides We construct a TM S to decide Theorem 4.9: is undecidable. Clearly, if R decides, then S decides. Because is undecidable, is undecidable too.

Automata & Formal Languages, Feodor F. Dragan, Kent State University 4 The Emptiness Problem for the Language of a TM. Theorem 5.2: is undecidable. Let assume that TM R decides. We construct a TM S to decide Idea is for S to run R on input and see whether it accepts. If it does then L(M) is empty and hence M does not accept w. But if M rejects …(???) we still do not know whether M accepts w. Instead of running R on we run R on a modification of ( ). The only string M1 accepts is w, so its language is nonempty if and only if it accepts w. S = “on input, an encoding of a TM M and a string w: 1.Use the description of M and w to construct the following TM M1. M1 = “on input x: 1.If, reject. 2.If x = w, run M on input w and accept if M does.” 2.Run R on input. 3.If R accepts, reject; if R rejects, accept.” The test whether x = w is obvious; scan the input and compare it character by character with w to determine whether they are the same. Note that S must be able to compute a description of M1 from a description of M and w. It is able because it needs only add extra states to M that perform the x = w test. If R were a decider for, S would be a decider for which is impossible.

Automata & Formal Languages, Feodor F. Dragan, Kent State University 5 The Equivalence Problem for TMs. Theorem 5.4: is undecidable. We could prove it by a reduction from, but we use this opportunity to give an example of an undecidability proof by reduction from Let TM R decides and construct TM S to decide as follows. S = “on input, an encoding of a TM M: 1.Run R on input, where M1 is a TM that rejects all inputs. 2.If R accepts, accept; if R rejects, reject.” The problem is a special case of the problem wherein one of the machines is fixed to recognize the empty language. This idea makes giving the reduction easy. So, If R were a decider for, S would be a decider for, which is impossible. One can also show that is neither Turing-recognizable nor co-Turing-recognizable. Read also page 183 where a simple problem called Post Correspondence Problem is shown to be unsolvable by algorithms.